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Uncertainty, page 3 CSI 4106, Winter 2005 Examples of approximate reasoning in real life Judging by general shape, not by details. Jumping to conclusions without sufficient evidence. (Consider a lawn sign, seen from afar.) Understanding language....

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Uncertainty, page 5 CSI 4106, Winter 2005 The Closed-World Assumption A complete theory in first-order logic must include either a fact or its negation. The Closed World Assumption (CWA) states that the only true facts are those that are explicitly listed as true (in a knowledge base, in a database) or are provably true. We may extend the theory by adding to it explicit negations of facts that cannot be proven. In Prolog, this principle is implemented by "finite failure", or "negation as failure".

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Uncertainty, page 6 CSI 4106, Winter 2005 Predicate completion The fact p(a) could be rewritten equivalently as ∀ x x = a  p(x) A completion of this quantified formula is a formula in which we enumerate all objects with property p. In this tiny example: ∀ x p(x)  x = a A larger example: the world of birds. ∀ x ostrich(x)  bird(x) ¬ostrich(Sam) bird(Tweety) To complete the predicate bird, we say: ∀ x bird(x)  (ostrich(x)  x = Tweety) This allows us to prove, for example, that ¬bird(Sam)

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Uncertainty, page 12 CSI 4106, Winter 2005 Taxonomic hierarchies and defaults (5) The last formula is equivalent to flying-ostrich(x)  false which reflects the fact that no taxonomical rule has flying-ostrich as a conclusion. We can now prove all of these: ¬flying-ostrich(Tweety) ¬ostrich(Tweety) ¬bird(Tweety) ¬abnormal t (Tweety) so we can show that ¬flies(Tweety)

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Uncertainty, page 13 CSI 4106, Winter 2005 Taxonomic hierarchies and defaults (6) This hierarchy again can be changed non-monotonically. Suppose that we add bird(Tweety) to the taxonomy. The new completion of the predicate bird will be: bird(x)  ostrich(x)  x = Tweety instead of bird(x)  ostrich(x) We will not be able to prove ¬abnormal t (Tweety) any more. We will, however, be able to prove ¬abnormal b (Tweety)

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Uncertainty, page 14 CSI 4106, Winter 2005 A quaker and a republican Quakers are pacifists. Republicans are not pacifists. Richard is a republican and a quaker. Is he a pacifist? These rules are ambiguous. Let us clarify: Only a typical quaker is a pacifist. Only a typical republican is not a pacifist. This can be expressed in terms of consistency: ∀ x quaker(x)  CONSISTENT(pacifist(x))  pacifist(x) ∀ x republican(x)  CONSISTENT(¬pacifist(x))  ¬pacifist(x) If we apply the first rule to Richard, we find he is a pacifist (nothing contradicts this conclusion), but then the second rule cannot be used—and vice versa. In effect, neither pacifist(x) nor ¬pacifist(x) can be proven.

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Uncertainty, page 15 CSI 4106, Winter 2005 Abduction... demonstrated on one example Abduction means systematic guessing: "infer" an assumption from a conclusion. For example, the following formula: ∀ x rainedOn(x)  wet(x) could be used "backwards" with a specific x : if wet(Tree) then rainedOn(Tree) This, however, would not be logically justified. We could say: wet(Tree)  CONSISTENT(rainedOn(Tree))  rainedOn(Tree) We could also attach probabilities, for example like this: wet(Tree)  rainedOn(Tree)|| 70% wet(Tree)  morningDewOn(Tree)|| 20% wet(Tree)  sprinkled(Tree)|| 10%

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Uncertainty, page 22 CSI 4106, Winter 2005 Bayes’s theorem Bayesian probability theory. Fuzzy logic (only signalled here). Dempster-Shafer theory (not discussed here). Bayes's theorem allows us to compute how probable it is that a hypothesis H i follows from a piece of evidence E (for example, from a symptom or a measurement). The required data: the probability of H j and the probability of E given H j for all possible hypotheses.

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Uncertainty, page 23 CSI 4106, Winter 2005 Bayes’s theorem (2) Medical diagnosis is a handy example. A patient may have a cold, a flu, pneumonia, rheumatism, and so on. The usual symptoms are high fever, short breath, runny nose, and so on. We need the probabilities (based on statistical data?) of all diseases, and the probabilities of high fever, short breath, runny nose in the case of a cold, a flu, pneumonia, rheumatism. [This is asking a lot!] We would also like to assume that all relationships between H j and E are mutually independent. [This is asking even more!]

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Uncertainty, page 30 CSI 4106, Winter 2005 Odds calculation (4) 25% of students in the AI course get an A. 80% of students who get an A do all homework. 60% of students who do not get an A do all homework. 75% of students who get an A are CS majors. 50% of students who do not get an A are CS majors. Irene does all her homework is the AI course. Mary is a CS major and does all her homework. What are Irene's and Mary's odds of getting an A? Let A = "gets an A". C = "is a CS major". W = "does all homework". Example